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Conditional Identities
Consider the equation, 2sin x 1  0 . We call equations like this conditional identities
because they are true only on the condition that x takes certain values. In this lecture, we learn to
solve for these conditional x-values.
Let’s consider the given equation, 2sin x 1  0 . Isolating the trigonometric expression,
we obtain, sin x  1 2 ; hence, x   6 and x  5 6 if we restrict x to values less than 2 .
Abdicating this restriction, the solutions reoccur every cycle of sine’s period be it in the positive
or negative direction. Consequently, we can say, 2sin x 1  0 if x   6  2k and
x  5 6  2k where k is an integer.
To understand our general condition above, let’s look at the solution x   6 . It is easy
to see that the identity is true again if x  13 6 (note that  6 and 13 6 are co-terminal
angles). Consider the expression x   6  2k where k = 1:
x
x
x
x

6

6

6

 2 k
 2 1 
 2

6
13
x
6
12
6
In the example above, we isolated the trigonometric expression. Another strategy
requires factoring. Consider the conditional identity: cot x cos2 x  2cot x . To solve for the
conditional values of x, we will set one side equal to zero and factor as below.
cot x cos 2 x  2 cot x
cot x cos 2 x  2 cot x  0
cot x  cos 2  2   0
We have a product equal to zero; hence, one of the factors must equal zero for the condition to be
true. Setting each factor equal to zero, we obtain solutions.
cot x  0
x

2
cos 2  2  0
 k
cos x   2
Note that one of the factors could not equal zero (  2 falls outside the range of cos x ).
Conditional identities can take the form of a quadratic as below.
2sin 2 x 1  sin x
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In which case, we apply strategies similar to a quadratic equation, i.e., we factor if possible and
otherwise rely on the quadratic formula.
2sin 2 x  1  sin x
2sin 2 x  sin x  1  0
 2sin x  1 sin x  1  0
2sin x  1  0, sin x  1  0
sin x  1 2,
sin x  1
7

 2 k , x   2 k 
6
2
11
x
 2 k
6
x
Sometimes using non-conditional trigonometric identities help solve a conditional
identity as below.
2sin 2 x  3cos x  3
2sin 2 x  3cos x  3
2 1  cos 2 x   3cos x  3
2  2 cos 2 x  3cos x  3  0
2 cos 2 x  3cos x  1  0
2 cos 2 x  3cos x  1  0
 2 cos x  1 cos x  1  0
2 cos x  1  0,
cos x  1  0
cos x  1 2,
cos x  1

 2k , x  2 k
3
5
x
 2 k
3
x
Sometimes squaring both sides offers an avenue towards a solution, but we must
remember that squaring both sides of an equation can introduce extraneous roots.
cos x  1  sin x
 cos x  1
2
  sin x 
2
cos 2 x  2 cos x  1  sin 2 x
cos 2 x  2 cos x  1  1  cos 2 x
2 cos 2 x  2 cos x  0
 cos x  1  2 cos x  0
cos x  1 or
x 
cos x  0
or x 
 3
2
,
2
Checking the solutions shows that 3 2 is extraneous, so the solutions are x    2 k and
x   2  2 k .
Some equations involve multiple angles. Start by solving for the multiple angle.
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2 cos  3 x   1  0
2 cos  3 x   1
cos  3 x  
1
2
Concentrating on the argument, 3x, we see:
3x 

3
 2 k and 3x 
5
 2 k
3
Dividing by the multiple, we obtain the solutions.
x

9

2 k
5 2 k
and x 
.

3
9
3
All the previous examples involve common angles (or some simple fraction of a common
angle). The solutions in the following conditional identity are not nice, convenient angles.
sec 2 x  2 tan x  4
1  tan 2 x  2 tan x  4
tan 2 x  2 tan x  3  0
 tan x  3 tan x  1  0
tan x  3  0,
tan x  1  0
tan x  3,
tan x  1
Here we come to an impasse, until we realize that we can undo a trigonometric function by its
inverse as demonstrated below.
arctan  tan x   arctan  3
arctan  tan x   arctan  1
x  arctan  3
x  arctan  1
x  arctan  3   k
x  arctan  1   k
Use a calculator to acquire approximations for the x-values above.
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Suggested Homework in Dugopolski
Section 6.6: #1-17 odd, #31-53 odd, #65, #69, #73, #75, #77, #79, #81, #85, #91, #93
Suggested Homework in Ratti & McWaters
Section 6.5: #7-25 odd, #39-45 odd, #55-69 odd
Section 6.6: #7-19 odd
Application Exercise
The distance d traveled by a projectile fired at an angle  is related to the initial
velocity v0 (in feet per second) by the equation v02 sin 2  32d . If the muzzle
velocity for a rifle equals 230 feet per second, then at what angle would it have to be
aimed for the bullet to travel 826.5625 feet?
Homework Problems
Find all the solutions on the interval 0  x  2π to the conditional identities.
#1 2sin  x   1  2
#2 2sin 2 x  sin x  0
#3 tan  2 x    3
#4 8cos  x  π4   6  10
#5 1  sin x  3 cos x
#6 cos 2 x  cos x  0
Find all the solutions to the conditional identities.
#7 cos θ  2sin θ
#8 sin 2θ  sin θ
#9 2sin 2 θ  9sin θ  5
#10 2sin 2 θ  3sin θ  1  0